Question: $\overline{AB} = \sqrt{65}$ $\overline{AC} = {?}$ $A$ $C$ $B$ $\sqrt{65}$ $?$ $ \sin( \angle BAC ) = \frac{4\sqrt{65} }{65}, \cos( \angle BAC ) = \frac{7\sqrt{65} }{65}, \tan( \angle BAC ) = \dfrac{4}{7}$
Solution: $\overline{AB}$ is the hypotenuse $\overline{AC}$ is adjacent to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the adjacent side so we can use the cos function (CAH) $ \cos( \angle BAC ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\overline{AC}}{\overline{AB}}= \frac{\overline{AC}}{\sqrt{65}} $ $ \overline{AC}=\sqrt{65} \cdot \cos( \angle BAC ) = \sqrt{65} \cdot \frac{7\sqrt{65} }{65} = 7$